Second-Order Differential Form 📂Geometry

Second-Order Differential Form

Overview

We define the binary operation $\wedge$ and, in the sense that we defined the first-order differential form, we define a second-order form for the differential manifold $M$ .

If differential manifolds seem difficult, one can think of them as $M = \mathbb{R}^{n}$ .

Buildup¹

Let’s consider the first-order form $\omega$ .

$\begin{align*} \omega : M &\to T^{\ast}M \\ p &\mapsto \omega_{p} \end{align*}$

This maps a point $p$ of a $n$ -dimensional differential manifold $M$ to an element $\omega_{p} \in T_{p}^{\ast}M$ of the cotangent space. Then, since $\omega_{p}$ is an element of the dual space of $T_{p}M$ , it is a functional as follows.

$w_{p} : T_{p}M \to \mathbb{R}$

In summary, the ‘first-order’ form can be thought of as mapping the point $p$ to a function $\omega_{p}$ taking ‘one’ tangent vector at $p$ as a variable. In this sense, we will define the ‘second-order’ form.

Wedge Product

Let’s call the function $\varphi : T_{p}M \times T_{p}M \to \mathbb{R}$ a bilinear alternating function.

$\varphi (v_{1}, v_{2}) = - \varphi (v_{2}, v_{1}),\quad v_{i} \in T_{p}M$

Let’s denote the set of such $\varphi$ s as $\Lambda^{2} (T_{p}^{\ast}M)$ .

$\Lambda^{2} (T_{p}^{\ast}M) := \left\{ \varphi : T_{p}M \times T_{p}M \to \mathbb{R}\ | \ \varphi \text{ is bilinear and alternate} \right\}$

Now, let’s define a binary operation $\wedge : T_{p}^{\ast}M \times T_{p}^{\ast}M \to \Lambda^{2} (T_{p}^{\ast}M)$ that sends two elements of $T_{p}^{\ast}M$ to $\Lambda^{2} (T_{p}^{\ast}M)$ . This means to express the elements of $\Lambda^{2} (T_{p}^{\ast}M)$ in terms of elements of $T_{p}^{\ast}M$ . Then, when we say $\varphi_{1}, \varphi_{2} \in T_{p}^{\ast}M$ , since $\Lambda^{2} (T_{p}^{\ast}M)$ is a set of alternating functions, the following must be true. (Note that the symbol $\wedge$ itself is read as [wedge], and the binary operation $\wedge$ is called the wedge product or exterior product. The TeX code is \wedge)

$\varphi_{1} \wedge \varphi_{2} \in \Lambda^{2} (T_{p}^{\ast}M)$

$(\varphi_{1} \wedge \varphi_{2}) (v_{1}, v_{2}) = - (\varphi_{1} \wedge \varphi_{2}) (v_{2}, v_{1}),\quad v_{i} \in T_{p}M$

If we define $\wedge$ as follows, it exactly satisfies the above condition.

$(\varphi_{1} \wedge \varphi_{2})(v_{1}, v_{2}) := \det \left[ \phi_{i}(v_{j}) \right]$

Here, $i$ represents the row index, and $j$ represents the column index. Of course, the wedge product $\wedge$ itself becomes an alternating function.

Alternating Property

$\begin{align*} (\varphi_{1} \wedge \varphi_{2})(v_{1}, v_{2}) =&\ \det \left[ \varphi_{i}(v_{j}) \right] \\ =&\ \begin{vmatrix} \varphi_{1}(v_{1}) & \varphi_{1}(v_{2}) \\ \varphi_{2}(v_{1}) & \varphi_{2}(v_{2}) \end{vmatrix} \\ =&\ - \begin{vmatrix} \varphi_{1}(v_{2}) & \varphi_{1}(v_{1}) \\ \varphi_{2}(v_{2}) & \varphi_{2}(v_{1}) \end{vmatrix} & \text{by property of determinant} \\ =&\ - (\varphi_{1} \wedge \varphi_{2})(v_{2}, v_{1}) \end{align*}$

■

Linearity

$\text{for } a\in \mathbb{R}$ ,

$\begin{align*} & (\varphi_{1} \wedge \varphi_{2})(av_{1} + v_{2}, w) \\[1em] =&\ \begin{vmatrix} \varphi_{1}(av_{1}+v_{2}) & \varphi_{1}(w) \\ \varphi_{2}(av_{1}+v_{2}) & \varphi_{2}(w) \end{vmatrix} \\[1em] =&\ \begin{vmatrix} a\varphi_{1}(v_{1}) + \varphi_{1}(v_{2}) & \varphi_{1}(w) \\ a\varphi_{2}(v_{1}) + \varphi_{2}(v_{2}) & \varphi_{2}(w) \end{vmatrix} & \text{by linearity of } \varphi_{i} \\[1em] =&\ \begin{vmatrix} a\varphi_{1}(v_{1}) & \varphi_{1}(w) \\ a\varphi_{2}(v_{1}) & \varphi_{2}(w) \end{vmatrix} + \begin{vmatrix}\varphi_{1}(v_{2}) & \varphi_{1}(w) \\ \varphi_{2}(v_{2}) & \varphi_{2}(w) \end{vmatrix} & \text{by property of determinant} \\[1em] =&\ a\begin{vmatrix} \varphi_{1}(v_{1}) & \varphi_{1}(w) \\ \varphi_{2}(v_{1}) & \varphi_{2}(w) \end{vmatrix} + \begin{vmatrix} \varphi_{1}(v_{2}) & \varphi_{1}(w) \\ \varphi_{2}(v_{2}) & \varphi_{2}(w) \end{vmatrix} & \text{by property of determinant} \\[1em] =&\ a(\varphi_{1} \wedge \varphi_{2})(v_{1}, w) + (\varphi_{1} \wedge \varphi_{2})(v_{2}, w) \end{align*}$

■

Basis

Now, let’s consider the wedge products of the basis $\left\{ (dx_{j})_{p} \right\}_{j}$ s of $T_{p}^{\ast}M$ . If you are quick to catch on, you might guess that these will be the basis of $\Lambda^{2} (T_{p}^{\ast}M)$ . For convenience, let’s denote it as follows.

$(dx_{i} \wedge dx_{j})_{p} \overset{\text{notation}}{=} (dx_{i})_{p} \wedge (dx_{j})_{p} \in \Lambda^{2} (T_{p}^{\ast}M)$

Then, $\left\{ (dx_{i} \wedge dx_{j})_{p} : i \lt j \right\}$ actually becomes the basis of $\Lambda^{2} (T_{p}^{\ast}M)$ , and the following holds.

$(dx_{i} \wedge dx_{j})_{p} = - (dx_{j} \wedge dx_{i})_{p},\quad i \ne j \\[1em] (dx_{i} \wedge dx_{i})_{p} = 0$

Now we are ready to define the second-order form.

Definition

We define the function $\omega : M \to \Lambda^{2} (T_{p}^{\ast}M)$ that maps a point $p \in M$ as follows as a second-order form in $M$ .

$\omega (p) = a_{12}(p)(dx_{1} \wedge dx_{2})_{p} + a_{13}(p)(dx_{1} \wedge dx_{3})_{p} + a_{23}(p)(dx_{2} \wedge dx_{3})_{p}$

$\omega$ is simply denoted as follows.

$\begin{align*} \omega =&\ a_{12}dx_{1} \wedge dx_{2} + a_{13}dx_{1} \wedge dx_{3} + a_{23}dx_{2} \wedge dx_{3} \\ =&\ a_{ij}dx_{i}\wedge dx_{j} (i \lt j) & \text{by Einstein notation} \end{align*}$

At this time, $a_{ij} : M \to \mathbb{R}$ . If each of $a_{ij}$ is differentiable, $\omega$ is called a second-order differential form.